Given that
\(\vec a, \vec b\) and \(\vec c\)are unit vectors.
i.e., |\(\vec a\)| = |\(\vec b\)| = |\(\vec c\)| = 1
And \(\vec a\) + \(\vec b\) + \(\vec c\) = \(\vec 0\)
Then \((\vec a+\vec b+\vec c).(\vec a+\vec b+\vec c)=\vec 0.\vec 0=0\)
⇒ \(\vec a\).\(\vec a\) + \(\vec b\).\(\vec b\) +\(\vec c\).\(\vec c\) + 2(\(\vec a\).\(\vec b\) +\(\vec b\).\(\vec c\) + \(\vec c\). \(\vec a\)) = 0
⇒ \(\vec a\).\(\vec b\) + \(\vec b\).\(\vec c\) + \(\vec c\).\(\vec a\) = -3/2 (\(\because \vec a.\vec a=|\vec a|^2=1
\)))
\(\therefore\vec a,\vec b\)and \(\vec c\) are not parallel to each other.
Now,
\(\vec a\times\vec b=\vec a\times(-(\vec a+\vec c))\) \((\because\vec b=-(\vec a+\vec c))\)
\(=-\vec a\times\vec a-\vec a\times\vec c\)
\(=-\vec a\times\vec c\) \((\because\vec a\times\vec a=\vec v)\)
\(=\vec c\times\vec a\)
And \(\vec b\times\vec c=-(\vec a+\vec c)\times\vec c\)
\(=-\vec a\times\vec c-\vec c\times\vec c\)
\(=\vec c\times\vec a(\because\vec c\times\vec c=\vec 0)\)
\(\therefore\vec a\times\vec b=\vec c\times\vec a=\vec b\times\vec c\neq\vec 0\)
(Because \(\vec a\),\(\vec b\) and \(\vec c\) are not mutually parallel)
Option (b) is correct